A small summary for ODE An Algebraic Approach (1) (2) (3)

In ODE, An Algebraic Approach (1)，

We factor the differential operator polynomial $D^n+a_{n-1}{D}^{n-1}+...+a_{1}D+a_0$$D^n+a_{n-1}D^{n-1}+...+ a_1D+a_0$

But in general, we do not know how to factor that. Thus we need Galois theory - Wikipedia to consider the solvability of $P(D)$$P(D)$

And we can observe that $(D+\lambda )=e^{-\lambda x}\circ D\circ e^{\lambda x}$$(D+\lambda)= e^{-\lambda x}\circ D\circ e^{\lambda x}$

That is amazing! If you are familiar with linear algebra, then you can remember that the spectral theorem

It looks like $A=P^{-1}BP$$A=P^{-1} B P$, see, $e^{\lambda x}$$e^{\lambda x}$ is eigenvector of $D$$D$

Amazing!

In ODE, An Algebraic Approach (2),

We talk about how to use formal power series to find the inverse of $P(D)$$P(D)$

We need the formal power series of $(D-\lambda )$$(D-\lambda)$ convergence,

thus we consider some spaces like ${\mathbb{P}}_n(x),\lambda =0$$\mathbb P_n(x),\lambda=0$ and $e^{\lambda x}{\mathbb{P}}_n(x)$$e^{\lambda x}\mathbb P_n(x)$

Because we know that $D$$D$ is nilpotent when it acts on ${\mathbb{P}}_n(x)$$\mathbb P_n(x)$

And $(D-\lambda )=e^{\lambda x}\circ D\circ e^{-\lambda x}$$(D-\lambda)= e^{\lambda x}\circ D\circ e^{-\lambda x}$, thus $(D-\lambda )$$(D-\lambda)$ is also nilpotent when it acts on $e^{\lambda x}{\mathbb{P}}_n(x)$$e^{\lambda x}\mathbb P_n(x)$

But, what if $\begin{array}{c}\frac{1}{P(D)}=a_0+a_{1}D+a_2{D}^2+...\end{array}$$\frac{1}{P(D)}=a_0+a_1D+a_2D^2+...\\$

Act on $g$$g$ is convergence although $g$$g$ is not a polynomial?

That is $\begin{array}{c}\underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^n{a}_i{g}^i\end{array}$$\lim_{n\to\infty}\sum_{i=0}^n a_i g^{i}\\$ exists

Which kind of function has this property?

In ODE, An Algebraic Approach (3),

You can see that, in fact, ODE (3) cover ODE (2)

The Method can solve more general ODE!

But in some cases, the method in ODE (2) is much more convenient.